How to draw lewis structure for any compound (neutral molecule or polyatomic ion) which contains only Main Group elements joined by two-atom shared-pair covalent bonds.
Step 1: Connectivity - arrange the atomic symbols so that covalently bonded atoms are contiguous.
# the number of covalent bonds an atom forms is called its valence.
# Some atoms have fixed valence. For example: H = 1, C = 4, F = 1.
# Some atoms have variable valence. For example: O = 2 (sometimes 3), B, N = 3 (sometimes 4).
# an atom bonded to only one other atom is peripheral (monovalent atoms such as H and F are always peripheral).
# an atom bonded to two or more other atoms is central.
# usually the formula is written to indicate connectivity. For example:
a)HCN => H bonded to C, C bonded to N, H and N are not bonded.
b)CH3OCH3 => three H bonded to C1, C1 bonded to O, O bonded to C2, C2 bonded to three H.
c)CH3CH2OH => three H bonded to C1, C1 bonded to C2, C2 bonded to two H and O, O bonded to H.
Otherwise, the least electronegative elements (if not monovalent) are central, the most electronegative elements are peripheral. For example:
CO3 2- = 3 peripheral O bonded to central C
PO4 3- = 4 peripheral O bonded to central P
H3O+ = 3 monovalent (therefore peripheral) H bonded to central O
The final Lewis Structure represents only the approx. electronic disribution,nt the actual 3-d arrangement of the atoms.
Step 2: Count the total number of Valence Shell Electrons nd divide them into pairs.
#.Sum the number of valence shell electrons of each atom;
# the number of valence shell electrons for an atom is equal to its group number # Subtract the charge on the assembly
The total should be even (these rules do no apply to assemblies with an odd number of electrons)
Examples:
HCN = 1+4+5-0 = 10 VSE = 5 VSE pairs
C2H6O = 2(4)+6(1)+6-0 = 20 VSE = 10 VSE pairs
CO3 2- = 4+3(6)-(-2) = 24 VSE = 12 VSE pairs
PO4 3- = 5+4(6)-(-3) = 32 VSE = 16 VSE pairs
H3O+ = 3(1)+6-(+1) = 8 VSE = 4 VSE pairs
Step 3: Assign Bond Pairs - Place one Valence Shell Electron pair of electrons between each bonded pair of atoms.
Connect each contiguous pair of atoms with one of the VSE pairs; each Bond Pair is shown as a line. Examples:
HCN = 5 VSE pairs - 2 BP = 3 pairs remaining
C2H6O = 10 VSE pairs - 8 BP = 2 pairs remaining
CO3 2- = 12 VSE pairs - 3 BP = 9 pairs remaining
PO4 3- = 16 VSE pairs - 4 BP = 12 pairs remaining
H3O+ = 4 VSE pairs - 3 BP = 1 pair remaining
Step 4: Assign Peripheral LP - Place up to three VSE pairs on each peripheral atom.
Distribute the VSE pairs remaining after Step 3 among the peripheral atoms as Lone Pairs (H cannot accept lone pairs; see the Rule of Orbitals below). At this stage, no peripheral atom may have more than 4 VSE pairs (1 BP + 3 LP). Examples
HCN : of 3 VSE pairs, all are assigned to N; no VSE pairs remain.
C2H6O : of 2 VSE pairs, none are assigned (since all peripheral atoms are H); 2 VSE pairs remain.
CO32- : of 9 VSE pairs, 3 are assigned to each O; no VSE pairs remain.
PO43- : of 12 VSE pairs, 3 are assigned to each O; no VSE pairs remain.
H3O+ : 1 pair, not assigned (since all peripheral atoms are H); 1 VSE pair remains.
Step 5: Assign Central LP - Place any remaining VSE pairs as Lone Pairs on central atom(s) according to the Rule of Orbitals.
The Rule of Orbitals: the total number of lone pairs and bond pairs (LP+BP) associated with an atom cannot exceed the number of Valence Shell Orbitals (VSO = n2, where n is the row of the Periodic Table in which that atom resides).
n = 1 (H): maximum VSE pairs (LP+BP) = VSO = 1;
n = 2 (B, C, N, O, F): maximum VSE pairs (LP+BP) = VSO = 4 ("octet rule")
n = 3 ((Al, Si, P, S, Cl): maximum VSE pairs (LP+BP) = VSO = 9; etc.
Examples
C2H6O = 2 pairs, both assigned to O since each C already has 4 BP.
H3O+ = 1 pair, assigned to O.
Step 6: Rearrange VSE Pairs - If necessary, push electron pairs according to the Rule of Orbitals and the Principle of Electroneutrality.
#Principle of Electroneutrality: each atom in a covalent molecular assembly has a formal charge close to zero.
#Formal Charge: FC = (Group Number) - (Bond Pairs) - 2(Lone Pairs)
#Electron Pushing: formally changing a lone pair into a bond pair, or vice versa, while retaining association with the atom.
Examples
HCN
Original Lewis Structure
H: FC = 1-1-2(0) = 0;
H: rule of orbitals satisfied (1 orbital, 1 VSE pair);
C: FC = 4-2-2(0) = +2;
C: rule of orbitals not satisfied (4 orbitals, only 2 VSE pairs; it must be associated with 4 VSE pairs);
N: FC = 5-1-2(3) = -2;
N: rule of orbitals satisfied (4 orbitals, 4 VSE pairs)
Rearranged Lewis Structure
Push two lone pairs on N into C-N bonding position, creating a C-N triple covalent bond.
H: FC = 0;
C: FC = 4-4-2(0) = 0;
N: FC = 5-3-2(1) = 0;
All atoms: rule of orbitals satisfied.
C2H6O (both isomers)
Original Lewis Structures
H (all): FC = 1-1-2(0) = 0;
H (all): rule of orbitals satisfied;
C (all): FC = 4-4-2(0) = 0;
C (all): rule of orbitals satisfied;
O: FC = 6-2-2(2) = 0;
O: rule of orbitals satisfied;
the original Lewis Diagrams (produced with rules 1-5) are correct, no rearrangement necessary.
CO32-
Original Lewis Structure
O (all): FC = 6-1-2(3) = -1;
O (all): rule of orbitals satisfied;
C: FC = 4-3-2(0) = +1;
C: rule of orbitals not satisfied; it must be associated with 4 VSE pairs.
Rearranged Lewis Structure
Push a lone pair on any of the three O into bonding position, creating one C-O double bond. For each of the three Lewis Structures thus produced:
O (single bond): FC = -1;
O (double bond): FC = 6-2-2(2) = 0
(average formal charge on each oxygen = -2/3);
C: FC = 4-4-2(0) = 0;
all atoms: rule of orbitals satisfied;
The set of three rearranged Lewis Structures correctly depicts the resonance in this ion.
PO43-
O (all): FC = 6-1-2(3) = -1;
O (all): rule of orbitals satisfied;
P: FC = 5-4-2(0) = +1;
P: rule of orbitals satisfied (P can accomodate up to 9 VSE pairs);
Push a lone pair on any of the four O into bonding position, creating one P-O double bond. For each of the four Lewis Diagrams thus produced:
O (single bond): FC = -1;
O (double bond): FC = 0
(average formal charge on each oxygen = -3/4);
P: FC = 5-5-2(0) = 0;
all atoms: rule of orbitals satisfied;
The set of four rearranged Lewis Diagrams correctly depicts the resonance structures of this ion.
H3O+
H (all): FC = 0;
H (all) rule of orbitals satisfied;
O: FC = 6-3-2(1) = +1;
O: rule of orbitals satisfied.
The original Lewis Structure is correct. The formal charge of +1 on oxygen is reduced by electronegative induction, so each H atom attains a slight positive charge while the charge on the O atom approaches 0.
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